# Contest problem without using heavy machinery in number theory

This is a contest problem (high school olympiad) a week ago:

Let $$f(x) = x^{4}+ 2x^{3} -2x^{2} - 4x + 4$$. Prove that there exists infinitely many prime $$p$$ such that $$f(m)$$ is not a multiple of $$p$$ for any integer $$m$$.

When I saw this problem, the first thing arose in my mind is to use Chebotarev's density theorem. If $$K$$ is a splitting field of $$f(x)$$ over $$\mathbb{Q}$$, then a prime $$p$$ satisfies the condition if and only if $$f(x)$$ doesn't have a solution in $$\mathbb{F}_{p}$$. According to SAGE, the splitting field of $$f(x)$$ has degree 8 and its Galois group is isomorphic to the Dihedral group $$D_{8}$$. By Chebotarev density theorem, the set of primes $$p$$ that inerts in $$K$$ has a positive density $$2/8 = 1/4$$, and $$f(x)$$ is irreducible over $$\mathbb{F}_{P}$$ for such prime $$p$$, which implies that it doesn't have a solution in $$\mathbb{F}_{p}$$.

I never prepared such high school contests seriously before, so this is the only solution I know until now. Is there any simple and elementary solution for this problem?

Edit: There are some other primes that satisfy the condition: if $$f(x)$$ factors as multiplication of two degree 2 polynomials, then $$f(x)$$ has no root in $$\mathbb{F}_{p}$$. The density of such primes is $$3/8$$. So actually $$5/8$$ of primes satisfy the condition.

• Not sure if this helps, but the expression can be 'factorised' into $x(x+2)(x^2-2)+4$. It is however indecomposable on $\mathbb Q[x]$. – YiFan Nov 14 '18 at 10:54

Note that $$f(x) = (x^2 + x - 2)^2 + x^2$$, and use the fact that if a prime $$p = 4k + 3$$ divides $$a^2 + b^2$$ then $$p$$ divides both $$a$$ and $$b$$.
• Thank you! How did you find such expression of $f(x)$? It seems very nontrivial for me. – Seewoo Lee Nov 14 '18 at 21:16
The FTA implies that every polynomial, in particular $$f(x)$$, can be factorised into the product of unique linear factors in $$\mathbb C[x]$$. Wolfram Alpha tells us that all of these factors have non-real coefficients. But any polynomial in $$\mathbb R[x]$$ can be factorised into the product of linear or quadratic factors, and since these aren't linear as we've just verified, it must be able to be factorised into quadratic factors. Two of them in fact, considering the degree of $$f$$. Looking at the leading coefficient it is clear that the factorisation can be expressed in the form $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ where both factors are indecomposable in $$\mathbb R[x]$$. Comparing coefficients: $$\begin{cases} a+c=2\\b+d+ac=-2\\ad+bc=-4\\bd=4 \end{cases}$$ which can be solved for $$(a,b,c,d)$$. The exact form is complicated but the point is that, if $$p\mid f(x)$$ then $$p\mid g(x)h(x)$$ (where $$g,h$$ are the factors in the previous factorisation). We already proved that $$g$$ and $$h$$ are indecomposable. Thus $$p\mid g(x)$$ and $$p\mid h(x)$$, which now implies that $$p\mid (a-c)x+(b-d)$$. Now from Wolfram Alpha we know that if we let $$\alpha=\sqrt{4+\sqrt{17}}$$ then $$a-c=2\alpha$$ and $$b-d=(3+\sqrt{17})\alpha$$, so we have $$2\alpha x+(3+\sqrt{17})\alpha$$. But $$x$$ is an integer and so both terms are irrational, hence their sum is irrational also. How can an integer divide a number which is not even rational? So we are done.
In the bolded statement above, just because $$g[x]$$ and $$h[x]$$ are indiscomposable in $$\mathbb R[x]$$ does not imply $$g(m)$$ and $$h(m)$$ are indiscomposable in $$\mathbb Z$$, which is why this is not a full solution. I would appreciate greatly if someone could complete this solution and fix this error.